Question
If a, b, c are in continued proportion, prove that
$\frac{a^2+b^2+c^2}{(a+b+c)^2}=\frac{a-b+c}{a+b+c}$

Answer

Given, a, b and c are in continued proportion.
Therefore,
$\frac{a}{b}$=$\frac{b}{c}$
$a c=b^2$
LHS $=\frac{a^2+b^2+c^2}{(a+b+c)^2}$
$=\frac{a^2+b^2+c^2+2 a b+2 b c+2 a c-(2 a b+2 b c+2 a c)}{(a+b+c)^2} $
$=\frac{(a+b+c)^2-2\left(a b+b c+b^2\right)}{(a+b+c)^2}$
$ =\frac{(a+b+c)^2-2 b(a+c+b)}{(a+b+c)^2} $
$=\frac{(a+b+c)(a+b+c-2 b)}{(a+b+c)^2} $
$ =\frac{a-b+c}{a+b+c}$
Hence proved.

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